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Lobsters In a study of the infestation of the Thenus orientalis lobster by two types of barnacles, Octolasmis tridens and \(O\). lowei, the carapace lengths (in millimeters) of 10 randomly selected lobsters caught in the seas near Singapore are measured: \(^{2}\) $$ \begin{array}{lllllllll} 78 & 66 & 65 & 63 & 60 & 60 & 58 & 56 & 52 & 50 \end{array} $$ Find a \(95 \%\) confidence interval for the mean carapace length of the \(T\). orientalis lobsters.

Short Answer

Expert verified
Answer: The 95% confidence interval for the mean carapace length of Thenus orientalis lobsters is (LL, UL), where you need to calculate LL and UL using the provided steps and data.

Step by step solution

01

Compute the sample mean and standard deviation

We will first calculate the sample mean and standard deviation for the given lengths: 78, 66, 65, 63, 60, 60, 58, 56, 52, 50 Sample mean (M) = (78+66+65+63+60+60+58+56+52+50) / 10 = 60.8 Sample standard deviation (s) can be calculated using the formula: $$s = \sqrt{\frac{1}{n-1}\sum_{i=1}^n{(x_i - \bar{x})^2}}$$ where n is the sample size and \(x_i\) are the individual lengths.
02

Find the t-value for the 95% confidence interval

For a 95% confidence interval and given sample size (n=10), we need to find the t-value based on the t-distribution with (n-1=9) degrees of freedom. Using the t-distribution table or calculator, the t-value for two-tailed test and 9 degrees of freedom at 95% confidence level is approximately 2.262.
03

Calculate the margin of error

The margin of error (MOE) can be calculated using the formula: $$MOE = t*s/\sqrt{n}$$ Plugging in the values we found: MOE = 2.262 * s / sqrt(10)
04

Compute the confidence interval

The confidence interval can be calculated as follows: Lower limit (LL) = M - MOE Upper limit (UL) = M + MOE We need to substitute M and MOE with the values found in previous steps.
05

Present the final result

The 95% confidence interval for the mean carapace length of Thenus orientalis lobsters is calculated to be: (LL, UL) Write in the calculated values once the standard deviation has been calculated from the data and use them in Step 3 and 4 to finalize the confidence interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the t-Distribution
A t-distribution is a type of probability distribution that is used when the sample size is small, usually less than 30, or when the population standard deviation is unknown. It's similar to the standard normal distribution but has thicker tails, which accounts for the extra variability that can occur in small samples.

This distribution is particularly handy in statistics for estimating the mean of a normally distributed population when the sample mean and standard deviation are known. In this exercise, because we have a sample size of 10, the t-distribution is more appropriate compared to the standard normal distribution, helping us account for the potential variability due to a smaller sample size. The area under the curve of the t-distribution helps us understand the level of significance, which is crucial when determining the t-value (as seen in Step 2 of the solution).

The t-distribution is indexed by degrees of freedom, which are related to the number of observations in the sample and is calculated as the number of observations minus one (n-1). For example, with our lobster study having 10 samples, we would use a t-distribution with 9 degrees of freedom. This helps ensure our calculations for confidence intervals are accurate and reliable.
What is a Sample Mean?
The sample mean is a way of finding the central tendency or the average of a set of numbers in a sample. In the context of our lobster lengths, it provides a single value that summarizes all 10 lobster carapace measurements. It is calculated by adding all individual values together and then dividing by the number of values.

Here, the carapace lengths add up to 608 mm and, when divided by 10, result in a mean of 60.8 mm. This sample mean is a crucial part of our calculations because it gives us an estimate of the average carapace length of a Thenus orientalis lobster in the population from which our sample was drawn.
  • The sample mean (\( \bar{x} \)) is used as a point estimate of the population mean.
  • In smaller samples, this mean might not exactly reflect the population mean, hence the need for confidence intervals.
The sample mean plays a vital role in constructing the confidence interval as it appears in many of the computation steps. It helps balance the data by considering extreme values which could otherwise misrepresent the general trend.
Explaining Standard Deviation
Standard deviation measures how spread out the numbers are in a data set. It tells us how much individual measurements deviate from the average (mean) measurement of the data set.

In our case of the lobster carapace lengths, it will let us know how varied these lengths are from the average length of 60.8 mm that we calculated.

Using the formula \( s = \sqrt{\frac{1}{n-1}\sum_{i=1}^n{(x_i - \bar{x})^2}} \), where \( n \) is the number of observations, you calculate the standard deviation by:
  • Calculating the difference between each value and the sample mean, \( \bar{x} \)
  • Squaring each difference
  • Summing all the squared differences
  • Dividing by \( n-1 \)
  • Taking the square root of the result
This calculated standard deviation helps us understand the variability in lobster lengths and is critical when calculating the margin of error. A smaller standard deviation indicates consistency in the lobster sizes, while a larger one suggests more variability.
Margin of Error in Context
The margin of error is the range within which we expect the true population parameter, like the mean carapace length in our lobster sample, to fall. It reflects the precision of our sample estimates and is an integral part of creating a confidence interval.

To compute the margin of error (MOE), we use the formula: \( MOE = t \cdot s / \sqrt{n} \), where \( t \) is the t-value at our desired confidence level, \( s \) is the standard deviation, and \( n \) is the number of observations.
  • The t-value factor adjusts for the sample size by reflecting how the confidence interval needs to be wider for smaller samples.
  • The standard deviation term ensures the interval reflects the spread or variability of the data.
In our study, if everything else stays constant, having a higher standard deviation would increase the margin of error, thus broadening our confidence interval. Likewise, a higher t-value (usually associated with smaller sample sizes or higher confidence levels) will also extend the margin of error.

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Most popular questions from this chapter

Refer to Exercise 10.94. Suppose that the word-association experiment is conducted using eight people as blocks and making a comparison of reaction times within each person; that is, each person is subjected to both stimuli in a random order. The reaction times (in seconds) for the experiment are as follows: $$ \begin{array}{ccc} \text { Person } & \text { Stimulus } 1 & \text { Stimulus 2 } \\ \hline 1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \\ 4 & 2 & 1 \\ 5 & 1 & 2 \\ 6 & 2 & 3 \\ 7 & 3 & 3 \\ 8 & 2 & 3 \end{array} $$ Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using \(\alpha=.05 .\)

Insects hovering in flight expend enormous amounts of energy for their size and weight. The data shown here were taken from a much larger body of data collected by T.M. Casey and colleagues. \({ }^{15}\) They show the wing stroke frequencies (in hertz) for two different species of bees, \(n_{1}=4\) Euglossa mandibularis Friese and \(n_{2}=6\) Euglossa imperialis Cockerell. $$ \begin{array}{cc} \text { E. mandibularis Friese } & \text { E. imperialis Cockerell } \\ \hline 235 & 180 \\ 225 & 169 \\ 190 & 180 \\ 188 & 185 \\ & 178 \\ & 182 \end{array} $$ a. Based on the observed ranges, do you think that a difference exists between the two population variances? b. Use an appropriate test to determine whether a difference exists. c. Explain why a Student's \(t\) -test with a pooled estimator \(s^{2}\) is unsuitable for comparing the mean wing stroke frequencies for the two species of bees.

Here are the red blood cell counts (in \(10^{6}\) cells per microliter) of a healthy person measured on each of 15 days: $$ \begin{array}{lllll} 5.4 & 5.2 & 5.0 & 5.2 & 5.5 \\ 5.3 & 5.4 & 5.2 & 5.1 & 5.3 \\ 5.3 & 4.9 & 5.4 & 5.2 & 5.2 \end{array} $$ Find a \(95 \%\) confidence interval estimate of \(\mu,\) the true mean red blood cell count for this person during the period of testing.

Under what assumptions can the \(F\) distribution be used in making inferences about the ratio of population variances?

As part of a larger pilot study, students at a Riverside, California middle school, will compare the learning of algebra by students using iPads versus students using the traditional algebra textbook with the same author and publisher. \(^{20}\) To remove teacher-to-teacher variation, the same teacher will teach both classes, and the iPad and textbook material are both provided by the same author and publisher. Suppose that after 1 month, 10 students were selected from each class and their scores on an algebra advancement test recorded. The summarized data follow. $$ \begin{array}{lcc} & \text { iPad } & \text { Textbook } \\ \hline \text { Mean } & 86.4 & 79.7 \\ \text { Standard Deviation } & 8.95 & 10.7 \\ \text { Sample Size } & 10 & 10 \end{array} $$ a. Use the summary data to test for a significant difference in advancement scores for the two groups using \(\alpha=.05 .\) b. Find a \(95 \%\) confidence interval for the difference in mean scores for the two groups. c. In light of parts a and b, what can we say about the efficacy of using an iPad versus a traditional textbook in learning algebra at the middle school level?

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